Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $a \neq 0$. $k = \dfrac{5a}{3(a + 8)} \times \dfrac{10(a + 8)}{-4} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $k = \dfrac{ 5a \times 10(a + 8) } { 3(a + 8) \times -4 } $ $ k = \dfrac{50a(a + 8)}{-12(a + 8)} $ We can cancel the $a + 8$ so long as $a + 8 \neq 0$ Therefore $a \neq -8$ $k = \dfrac{50a \cancel{(a + 8})}{-12 \cancel{(a + 8)}} = -\dfrac{50a}{12} = -\dfrac{25a}{6} $